Line Integral of Every Positively Oriented Simple Closed Path - Green's Theorem

Question

This question is from Example #5, Section 16.4 on P1059 of Calculus, 6th Ed, by James Stewart.

Given Question: If $\mathbf{F}(x,y) = \left(\dfrac{-y}{x^2 + y^2}, \dfrac{x}{x^2 + y^2}\right)$, show that $\int_C \mathbf{F} \cdot d\mathbf{r} = 2\pi $ for every positively oriented simple closed path that encloses the origin.

Part of the Given Solution: Since $C$ is an ARBITRARY closed path that encloses the origin, it's difficult to compute the given integral directly. So let's consider a counterclockwise circle $A$ with center the origin and radius $a$, where $a$ is chosen to be small enough that $A$ lies inside $C$, as indicated by the picture below. Let $D$ be the region bounded by $C$ and $A$.
Then positively oriented $\partial D = C \cup (-A)$. So the version of Green Theorem's applied to regions with holes gives:

$\int_C \mathbf{F} \cdot d\mathbf{r} + \int_{-A} \mathbf{F} \cdot d\mathbf{r} = \iint_D \underbrace{(\partial_x Q - \partial_y P)}_{\Large{= 0}} \, dA $ $\Longrightarrow \int_C \mathbf{F} \cdot d\mathbf{r} = \int_{A} \mathbf{F} \cdot d\mathbf{r}$.
(Rest of solution omitted)

$\Large{\text{Q1.}}$ I can't perceive how one would divine to construct $A$ to solve this problem. So I feel that my following solution is more intuitive. Is it really? However, I don't seem to need Green's Thoerem. How and where would it be applied?

My solution: It's hard to calculate $\int_C \mathbf{F} \cdot d\mathbf{r} $. The idea then is to find one $A$ so that
$(\dagger) \int_{A} \mathbf{F} \cdot d\mathbf{r}$ is easier to compute and
$(\ddagger)\int_C \mathbf{F} \cdot d\mathbf{r} = \int_{A} \mathbf{F} \cdot d\mathbf{r}$.

Now, $(\ddagger) \iff \int_C \mathbf{F} \cdot d\mathbf{r} \huge{\color{red}{-}} \normalsize\int_A \mathbf{F} \cdot d\mathbf{r} = 0 \iff \int_C \mathbf{F} \cdot d\mathbf{r} +\int_{\huge{\color{red}{-}\normalsize{A}}} \mathbf{F} \cdot d\mathbf{r} = 0 $ $\iff \int_{\Large{\partial D} \normalsize{ \, = \, (C} \, \cup \huge{\color{red}{-}\normalsize{A})}} \mathbf{F} \cdot d\mathbf{r} = 0$.

For $\partial D$ to be positively oriented, $C$ and $\color{red}{-}A$ must both be positively oriented $\Longrightarrow A$ must be NEGATIVELY oriented, so $D$ must be to the right of $A$. One convenient choice of $A$ would just be a counterclockwise circle with center the origin and the radius $a$.

$\Large{\text{Q2.}}$ Why don't I get the requested answer if I pick $A$ as a counterclockwise ellipse enclosing the origin? In other words, $A$ is parameterised via $r(t) = (a\cos t, b\sin t), 0 \leq t \leq 2\pi $.

Then $\int_{A} \mathbf{F} \cdot d\mathbf{r} = \int_0^{2 \pi} \left(\dfrac{-b\sin t}{a^2 \cos^2 t + b^2 \sin^2 t},\dfrac{a\cos t}{a^2 \cos^2 t + b^2 \sin^2 t} \right) \cdot (-a\sin t, b \cos t) \, dt$.
$ = ab \int_0^{2 \pi} \dfrac{1}{a^2 \cos^2 t + b^2 \sin^2 t} \, dt $.
If $a = b$, then the requested answer follows by inspection. But what if $a \neq b$?

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$\Large{\text{Supplementaries to Muphrid's Answer: }}$

$\Large{\text{Q1.1.}}$ To rephrase your answer, are you saying that my solution has to justify why there exists an $A \ni \int_C \mathbf{F} \cdot d\mathbf{r} = \int_{A} \mathbf{F} \cdot d\mathbf{r}. ?$ This is true because $\int_C \mathbf{F} \cdot d\mathbf{r} = \int_{A} \mathbf{F} \cdot d\mathbf{r} \iff \int_{\Large{\partial D} \normalsize{ \, = \, (C} \, \cup \huge{\color{red}{-}\normalsize{A})}} \mathbf{F} \cdot d\mathbf{r} = 0, $ where the last equation is true by Green's Theorem: $\int_{\Large{\partial D} \normalsize{ \, = \, (C} \, \cup \huge{\color{red}{-}\normalsize{A})}} \mathbf{F} \cdot d\mathbf{r} = \iint_D \underbrace{(\partial_x Q - \partial_y P)}_{\Large{= 0}} \, dA. $

$\Large{\text{Q1.2.}}$ I wrote that $\partial D \normalsize{ \, = \, (C} \, \cup \huge{\color{red}{-}\normalsize{A})} $. Are you saying that this is wrong and that $\partial D \normalsize{ \, \subsetneqq \, (C} \, \cup \huge{\color{red}{-}\normalsize{A})} $?

$\Large{\text{Q1.3.}}$ I understand that $\mathbf{F}(0,0)$ is undefined. However, how does this imply that integrals enclosing the origin pick up $2\pi$?

$\Large{\text{Q1.4.}}$ How is $(\nabla \times \mathbf{F}) \cdot d\mathbf{A} = \partial_x Q - \partial_y P $? How is $d\mathbf{A} = z \, dx \, dy ?$

$\Large{\text{Q2.1.}}$ I understand that using a circle works fine and is easier. I just want to derive the same answer with an ellipse though. What went wrong?

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$\Large{\text{Supplementaries to Muphrid's 2nd Comment: }}$

$\Large{\text{Q1.3.1.}}$ You write: "When you choose a unit circular path, the resulting integral looks like an arclength integral."

Are you saying that choice of a unit circular path $\Longrightarrow \int_{A} \mathbf{F} \cdot d\mathbf{r} = ... = \int_{0}^{2\pi} 1 \, dt = 2\pi? $ How are you so confident that $...$ always yields this answer?

$\Large{\text{Q1.4.1.}}$ Why can you define $d\mathbf{A} = z \, dx \, dy ?$ We're not given $d\mathbf{A}$?

Answer 1

1) First, while you've chosen the orientation of $A$ correctly, you should realize that the boundary $\partial D$ of the region $D$ contains $C$ and $-A$. This is important.

Now, apply Green's theorem:

$$\oint_{\partial D} F \cdot dr = \int_C F \cdot dr - \int_A F \cdot dr = \int_D (\nabla \times F) \cdot dA$$

You should argue that the area integral is zero. Instead, in your second point ($\ddagger$), you come to this conclusion but without the supporting logic to back it up (which must come from Green's theorem).

(For fun: realize that $\nabla \times F \neq 0$ at the origin, and this is why integrals enclosing the origin pick up $2\pi$.)

2) Why would you pick an ellipse instead of a circle? Green's theorem tells us that $C$ could be an ellipse instead of an arbitrary curve and thus the integral would have to be $2\pi$, but why put yourself through that work?

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Responses to supplemental questions:

1.1) Yes, that is the correct logic.

1.2) No, that is the correct expression for the boundary.

1.3) I'm just observing that the line integral enclosing the origin can only be nonzero if the curl is somewhere nonzero in the region enclosed. You can also understand this in relation to complex analysis, with the point at the origin being a pole, which has a nonzero residue.

1.4) Yes, I wrote this incorrectly. I've added the necessary cdot to the original formula.

2.1) It doesn't seem to me there's anything wrong with that integral; it's just not easy to solve (wolfram gives the right answer for specific values of $a$, $b$, but not symbolically). You could try, for example, the polar equation of an ellipse, but I can't guarantee that would be any easier.

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Responses to supplemental supplemental questions:

1.3.1) This problem is very specifically constructed so that $F = \hat \theta /r$. Note the hat: this denotes a unit vector. $\hat \theta$ is always tangent to curves of constant $r$ and is in the direction of increasing $\theta$. When you choose a unit circular path, $r = 1$, and $d\mathbf r/d\theta = r \hat \theta$ also, rendering the integral

$$\int_A F \cdot d\boldsymbol \ell = \int_0^{2\pi} \frac{\hat\theta}{r} \cdot \frac{\hat \theta}{r} \, d\theta = \int_0^{2\pi} \hat \theta \cdot \hat \theta \, d\theta$$

Since $\hat \theta \cdot \hat \theta = 1$, the form I argued follows. That is precisely why choosing a unit circular path is so convenient here.

1.4.1) You have the freedom to choose the surface whose boundary is a given curve. That choice of surface then determines the differential. I don't boldface vectors, so you'll have to figure out from context whether a quantity is scalar or vector. Nevertheless, I wrote $dA = \hat z \, dx \, dy$, and $\hat z$ is a unit vector. The choice of surface that goes with this is that of using the flat $xy$-plane, which is the most convenient (I dare say necessary, as you have no information about what $F$ does outside of the $xy$-plane).

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Edit: the definition of $\hat \theta$ is

$$\hat \theta = \frac{1}{r} \frac{\partial \mathbf r}{\partial \theta}$$

write $\mathbf r = (r \cos \theta, r \sin \theta)$, and you get

$$\hat \theta = \frac{1}{r} (-r \sin \theta, r \cos \theta) = \left(- \frac{y}{\sqrt{x^2 + y^2}}, \frac{x}{\sqrt{x^2+ y^2}} \right)$$

You should be able to see that $\hat \theta/r = F$ then.

Answer 2

You see the difficulty at the wrong place: Which orientation is chosen for the circle $A$ is irrelevant, as long as the correct signs are used in the subsequent argument.

The real problems are elsewhere:

(i) To which kinds of domains is Green's theorem applicable, i.e. for which $\Omega\subset{\mathbb R}^2$ is there a boundary cycle $\partial\Omega$ such that Green's theorem holds?

(ii) How do we know that a "simple closed curve" $\gamma\subset{\mathbb R}^2$ actually bounds some region $\Omega$, such that we can speak of "interior points"? This is Jordan's curve theorem, which has no simple proof.

All this is glossed over in the statement of your exercice.